What is the integral from 0 to 2 of pi(16-y^4) ?

The integral from 0 to 2 of pi(16-y^4) is pi 128/5

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integral from 0 to 2 of pi(16-y^4)

\int_{0}^{2} π (16 - y^{4}) d y

π \frac{128}{5}

Decimal Notation

80.42477 \dots

Solution steps

\int_{0}^{2} π (16 - y^{4}) d y

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= π \cdot \int_{0}^{2} 16 - y^{4} d y

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= π (\int_{0}^{2} 16 d y - \int_{0}^{2} y^{4} d y)

\int_{0}^{2} 16 d y = 32

\int_{0}^{2} y^{4} d y = \frac{32}{5}

= π (32 - \frac{32}{5})

Join

32 - \frac{32}{5} : \frac{128}{5}

= π \frac{128}{5}

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What is the integral from 0 to 2 of pi(16-y^4) ?
The integral from 0 to 2 of pi(16-y^4) is pi 128/5